Question

Let $f\left(x\right)=1+\underset{0}{\overset{1}{\int }}(x{e}^y+y{e}^x)f\left(y\right)dy$ where $x$ and $y$ are independent variables. If complete solution set of $x$ for which the function $h\left(x\right)=f\left(x\right)+3x$ is strictly increasing is $(-\infty ,k),$ and $[.]$ denotes the greatest integer function, then $\left[\frac{4}{3}{e}^k\right]$ equals to

• A. $12$
• B. $2$
• C. $3$
• D. $9$

Answer 1

The correct option is C $3$

$f\left(x\right)=1+x\underset{0}{\overset{1}{\int }}{e}^{y}f\left(y\right)dy+e^x\underset{0}{\overset{1}{\int }}yf\left(y\right)dy$
Let $A=\underset{0}{\overset{1}{\int }}{e}^{y}f\left(y\right)dy$ and $B=\underset{0}{\overset{1}{\int }}yf\left(y\right)dy$
Then, $f\left(x\right)=1+Ax+B{e}^x$

$A=\underset{0}{\overset{1}{\int }}{e}^y(1+Ay+B{e}^y)dy\Rightarrow A={[e^y+A(y{e}^y-e^y)+\frac{B{e}^{2y}}{2}]}_0^1\Rightarrow A=e+\frac{B{e}^2}{2}-(1-A+\frac{B}{2})\Rightarrow B=-\frac{2}{e+1}$

Now, $B=\underset{0}{\overset{1}{\int }}y(1+Ay+B{e}^y)dy$
$\Rightarrow B={[\frac{y^2}{2}+\frac{A{y}^3}{3}+B(y{e}^y-e^y\left)\right]}_0^1\Rightarrow B=\frac{1}{2}+\frac{A}{3}-(0+0-B)\Rightarrow A=-\frac{3}{2}$

Given that $h\left(x\right)=f\left(x\right)+3x$ is strictly increasing in $(-\infty ,k)$
$\Rightarrow f^{\prime }\left(x\right)+3>0$
$\Rightarrow -\frac{3}{2}-\frac{2e^x}{(e+1)}+3>0\Rightarrow e^x<\frac{3(e+1)}{4}$
$\therefore \left[\frac{4}{3}{e}^k\right]=[e+1]=3$